Evaluate cos(arcsin(0))

Draw a triangle in the plane with vertices ${\displaystyle (\sqrt{1^2-0^2},0)}$, ${\displaystyle (\sqrt{1^2-0^2},0)}$, and the origin. Then ${\displaystyle \arcsin \left(0\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (\sqrt{1^2-0^2},0)}$. Therefore, ${\displaystyle \cos \left(\arcsin \left(0\right)\right)}$ is ${\displaystyle \sqrt{1}}$.